Another Way of Factorising Cubics.
In the previous page, we had introduced a method, we termed the quickest to factorise cubics. Now we give you another method, which is also convenient.
Remember, cubics have the form:
$${f(x)=ax^3+bx^2+cx+d}$$Using the factor theorem, you will first need to find, \(f(x) = 0\).Then multiply it by some quadratic expression ( \(\,mx^2 +nx+p)\,\). The expression
Let us try a few examples:
-
Factorise \(f(x)=x^3-5x^2-8x+12\)
The steps:
- Find the factor, using factor theorem. This means try different values of \(x\) that will make \(f (\,x)\, =0\). In this case, \( x = 1\) will make \(f (\,x)\, =0\). The factor is therefore, \((\,x-1)\,\)
- Now we must multiply this factor by the quadratic.
$${f(\,x)\,=(\,x-1)\,(\,mx^2 +nx+p)\,}$$
-
The coefficient of the \(x^2\) of the quadratic equals the coefficient of \(x^3\). Therefore, \(m=1\).
-
Find the sum of \(x\) multiplied by \(nx\) and \(-1\) multiplied by \(mx^2\) and equate them to \(bx^2\) of the quadratic.
\begin{gather*} nx^2 + (-mx^2) = bx^2 \\ nx^2 -mx^2 = -5x^2 \\ nx^2 -x^2 = -5x^2 \\ nx^2 = -4x^2 \\ n = -4 \end{gather*} Multiply \(-1\) by \(p\) and equate it to \(d\).
\begin{gather*} -p=d\\ -p=12\\ p=-12 \end{gather*}
-
Now substitute the values in the expression
\begin{gather*} f(\,x)\,= x^3-5x^2-8x+12\\ f(\,x)\, = (\,x-1)\,(\,x^2-4x-12)\, \\ f(\,x)\, = (\,x-1)\,(\,x-6)\,(\,x+2)\, \end{gather*}
The expression has been factororised.
All these can be done in one step. But it gets better only with more practice.
-
Factorise \(f(x)=-x^3+10x^2-17x-28\)
The factor is \((\,x+1)\,\).
By following the steps in (1), we get:
\begin{gather} m=-1\\ n=11\\ p=-28 \end{gather}The expression becomes:
\begin{gather*} f(\,x)\,= -x^3+10x^2-17x-28\\ f(\,x)\, = (\,x+1)\,(\,-x^2+11x-28)\, \\ f(\,x)\, = (\,x+1)\,[\,-(\,x^2-11x+28)\,]\, \\ f(\,x)\, = (\,x+1)\,[\,-(\,x-7)\,(\,x-4)\,]\, \\ f(\,x)\, = -(\,x+1)\,(\,x-6)\,(\,x-4)\, \end{gather*}
If this method does not work for you, try another one here.
